Tham khảo tài liệu 'finite element analysis - thermomechanics of solids part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 2 Mathematical Foundations Tensors TENSORS We now consider two n X 1 vectors v and w and an n X n matrix A such that v Aw. We now make the important assumption that the underlying information in this relation is preserved under rotation. In particular simple manipulation furnishes that v Qv QAw M QAQTQw QAQTw . The square matrix A is now called a second-order tensor if and only if A QAQT. Let A and B be second-order n X n tensors. The manipulations that follow demonstrate that AT A B AB and A 1 are also tensors. At QAQt t QtTAtQt A B QAQt QBQt QABQt A B A B QAQt qbqt Q A B Qt A -1 QAQt -1 qt-1a-1q-1 QA-1Qt . 25 2003 by CRC CRC Press LLC 26 Finite Element Analysis Thermomechanics of Solids Let x denote an n X 1 vector. The outer product xxT is a second-order tensor since xxT x x T Qx Qx T Q xxT Qt Next d20 dxTHdx H fdi f x v dx J v dx J However dx tH dx Q dx TH Q dx dxT QTH Q dx from which we conclude that the Hessian H is a second-order tensor. Finally let u be a vector-valued function of x. Then du dt dx from which duT dxT fdu i vox J and also duT dxTf i uT. vdx J We conclude that f du i T vdx J duT dxT Furthermore if du is a vector generated from du by rotation in the opposite sense from the coordinate axes then du Qdu and dx Qdx . Hence Q is a tensor. Also since du 1 - dx it is apparent that dx 1 Q f QT dx dx from which we conclude that 1 is a tensor. We can similarly show that I and 0 are tensors. x 2003 by CRC CRC Press LLC Mathematical Foundations Tensors 27 DIVERGENCE CURL AND LAPLACIAN OF A TENSoR Suppose A is a tensor and b is an arbitrary spatially constant vector of compatible dimension. The divergence and curl of a vector have already been defined. For later purposes we need to extend the definition of the divergence and the curl to A. Divergence Recall the divergence theorem J cTndS J VTcdV. Let c ATb in which b is an arbitrary constant vector. Now bT J AndS Jvt ATb dV J VtAt dVb bT J .
